Question

Jason mixed 8 ml of a salt solution with 11ml of a 47% salt solution,to make a 43% salt solution. Find the percent salt concentration of the first solution

Answer 1

Let x% be the salt concentration of first solution.

Then in 8ml, amount of salt =  $\frac{8x}{100}$

Given that to this 8ml solution, 11ml solution of a 47% salt solution is mixed.

amount of salt in 11ml of 47% salt solution is  =  $\frac{11*47}{100} = \frac{517}{100}$

So, total amount of salt in new mixture =  $\frac{8x}{100} +\frac{517}{100} =\frac{8x+517}{100}$

But given that this new solution is a 43% salt solution.

Hence amount of salt in new mixture of (8+11)=19ml solution of 43% salt solution is $\frac{19*43}{100} = \frac{817}{100}$

Hence  $\frac{8x+517}{100} = \frac{817}{100}$

            8x+517 = 817

             8x=817-517 = 300

             x= 300/8 = 37.5

Hence the salt concentration in the first solution is 37.5%